Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises - Page 704: 44

Answer

Converges to $8$

Work Step by Step

Given: $a_n=\sqrt[n] {2^{1+3n}}$ It can be re-written as $a_n= ({2^{1+3n}})^{1/n}=2^(\frac{{1+3n}}{n})$ $\lim\limits_{n \to \infty}a_n=\lim\limits_{n \to \infty}2^(\frac{{1+3n}}{n})$ $=\lim\limits_{n \to \infty}2^{(\frac{1}{n}+\frac{{3n}}{n})}$ $=\lim\limits_{n \to \infty}2^{(\frac{1}{n}+3)}$ $=2^{(0+3)}$ $=2^{3}$ $=8$ Therefore, the given sequence converges to $8$.
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